Suppose we turn over cards simultaneously from two well shuffled deck of ordinary playing cards. We say we obtain an exact match on a particular turn if the same card appears from each deck. Find the probability of at least one match.
Hint:Let $C_i$ denote the event on exact match on the ith turn then P($C_i$) = 1/52 and P($C_i$ $\bigcap$ $C_j$) = 50!/52!.
Could someone please explain how to calculate P($C_i$) and P($C_i$ $\bigcap$ $C_j$).
Hint: The probability of an exact match on the $i$th turn is precisely the probability that, no matter what card was turned over in deck 1, the exact same card is turned over in deck 2. There are 52 ways to choose the card from deck 1; given the card from deck 1 there is only 1 way to choose a card from deck 2.
Getting exact matches in both $i$ and $j$: choose which card appears in deck 1 at position $i$ and at position $j$. (There are $52\cdot51$ ways to do this.) Now, there's only one right way to choose the cards in positions $i$ and $j$ in deck 2.
Let $C^{1} = c_{1}^{1},c_{2}^{1},...,c_{52}^{1}$ be the first deck of cards, and $C^{2} = c_{1}^{2},c_{2}^{2},...,c_{52}^{2}$ be the second deck of cards, as well. We know $C^{1}$ and $C^{2}$ are independent. Thus we have
$\Pr(C_{i}) = \Pr(c_{i}^{1} = c_{i}^{2}) = \sum_{j=1}^{52}\Pr(c_{i}^{1} = c_{i}^{2}=j) = \sum_{j=1}^{52}\Pr(c_{i}^{1} =j)\Pr(c_{i}^{2} =j) = 52\times \frac{1}{52} \times \frac{1}{52}=\frac{1}{52}$